Shankar, R. (1994), *Principles of Quantum Mechanics*, Plenum Press. Chapter 12, Exercise 12.2.4.

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When it comes to symmetries in quantum mechanics, we’ve looked at translations and rotations in two dimensions, and found that the generators are the momenta and for translations, and the angular momentum for rotations.

From the fact that does not commute with either momentum or position operators, you might guess that if we performed some sequence of translations and rotations on a system that the order in which these operations are done matters. In fact, you can see this by considering simple two-dimensional geometry, without reference to quantum mechanics. Consider the and axes on a sheet of graph paper. First, translate these axes by adding the vector to all points, so that the new origin of coordinates lies at position as referenced in the original coordinates. Next, do a rotation *about the original origin* by some angle . This will move the new origin around the original axis. Now, do the inverse of the original translation by adding to all points. Finally, do the inverse of the rotation by rotating the system by around the *original* axis. You’ll find that the axes that have undergone this sequence of transformations does not coincide with the original axes. However, if you did the same set of four transformations in the order: translate by , translate by , rotate by , rotate by , the transformed axes *would* coincide with the original axes.

To see how this works in quantum mechanics, we can again consider infinitesimal translations and rotations. If we start with a point at location and apply the four transformations described above, but now for an infinitesimal translation and rotation , then the successive transformations work as follows. In each case, we’ll retain terms up to order and but discard terms of order , , and higher. [I’m not quite sure of the rationale that allows us to do this, apart from the fact that it gives the right answer.]

Thus, to this order in the infinitesimals, the combination of translation-rotation-translation-rotation is equivalent to a single translation by a distance . We can write this in terms of the unitary quantum operators for translations and rotations as

Using the forms of these operators for infinitesimal transformations, we can expand both sides to give

Since the infinitesimal displacements are arbitrary, this equation can be valid only if the coefficients of each combination of and are equal on both sides. As above, we’ll discard any terms of order , , and higher. The algebra is straightforward although a bit tedious, so I’ll just give a couple of examples here.

The coefficient of on its own is, on the LHS

On the RHS, there is no term in , so we get 0 on the RHS. In this case, we see the equation is consistent.

For the term, we get on the LHS:

On the RHS, the term is

Thus the condition here becomes

which agrees with the commutation relation we found earlier. By considering the coefficient of , we arrive at the other condition, which is

The result of this calculation doesn’t tell us anything new about the translation or rotation operators, but it does show that the condition 7 is consistent with what we already know about the commutators of position, momentum and angular momentum.

As Shankar points out, we might think that we need to verify the conditions for an infinite number of combinations of rotations and translations, since each such combination gives rise to a different overall transformation. He says that it has actually been shown that the example above is sufficient to guarantee that all such combinations do in fact give valid results, although he doesn’t give the details. We are, however, given the exercise of verifying this claim for one special case, which we’ll consider now.

In this example, we’ll consider the same four transformations, in the same order, as above except that we’ll take the translation to be entirely in the direction so that . This time, we’ll retain terms up to and see what we get. We start by repeating the calculations in 1 through 6. However, because we’re saving higher order terms, we need to represent the infinitesimal rotations by

That is, we’re approximating by the first two terms in its expansion. Using this, we have

To get the last line, I used Maple to do the algebra in multiplying out the terms. At this point, we can neglect the terms in , leaving us with the overall transformation:

This is equivalent to a translation by , so by analogy with 7, we have the condition

To expand the operators on the LHS and retain terms up to , we need to expand the rotation operators up to order . Treating the rotation operator as an exponential, this expansion is

Using this approximation gives us

By equating the coefficients of we regain 12, so that condition checks out.

Extracting the coefficient of on the LHS gives

Matching this to the term on the RHS of 24, we get the condition specified in Shankar’s problem:

We can show that this condition reduces to the already-known commutators by using the identity

Applying this to 26 we have

Thus the more complicated condition 26 actually reduces to existing commutators.